Question

The largest area of a trapezium inscribed in a semicircle of radius $R$, if the lower base is on the diameter, is

• A. $\frac{3\sqrt{3}{R}^2}{4}$
• B. $\frac{3R^2}{\sqrt{2}}$
• C. $\frac{3\sqrt{3}{R}^2}{8}$
• D. $R^2$

Answer 1

The correct option is A

$\frac{3\sqrt{3}{R}^2}{4}$

Explanation for the correct option

Step 1: Find the area of trapezium inside the given semicircle.

Visualize the given situation as follows:

Here, $ABCD$ is a trapezium and assume that $\angle BAD=\theta$ and $DE$ and $CF$ are the perpendicular to the line $AB$ such that, $AE=BF$ and $DE=CF$.

We know that the angle in a semicircle is of $90°$.

So, $\angle ADB=90°$.

From the right-angle triangle $ADB$.

$AD=AB\cos \theta$

Since, $AB$ is the diameter of the given semicircle.

So, $AB=2R$.

Therefore, $AD=2R\cos \theta$.

From the right-angle triangle $AED$.

$AE=AD\cos \theta$

Since, $AD=2R\cos \theta$.

So, $AE=2R{\cos }^2\theta$

Since, $AB=AE+EF+FB$

So,

$$\begin{gathered} AB=2AE+EF \\ \Rightarrow EF=AB-2AE \\ \Rightarrow EF=2R-4R{\cos }^2\theta \end{gathered}$$

From the figure it is clear that, $EF=CD$.

Therefore, $CD=2R-4R{\cos }^2\theta ...\left[1\right]$.

Also, From the right-angle triangle $AED$.

$DE=AD\sin \theta$

Since, $AD=2R\cos \theta$.

So, $DE=2R\cos \theta \sin \theta ...\left[2\right]$

Now, the area $A$ of the trapezium can be given by: $\frac{1}{2}(AB+CD)DE$.

So, from equation $\left[1\right]$ and $\left[2\right]$.

$$\begin{gathered} A=\frac{1}{2}(2R+2R-4R\cos \theta )2R\cos \theta \sin \theta \\ \Rightarrow A=(4R-4R\cos {\theta }^2)R\cos \theta \sin \theta \\ \Rightarrow A=4R^2\cos \theta {\sin }^3\theta \end{gathered}$$

Step 2: Find the maximum area of trapezium inside the given semicircle.

Since, the area of the trapezium in the given semicircle is $A=4R^2\cos \theta {\sin }^3\theta$.

Differentiate both sides with respect to $\theta$.

$$\begin{gathered} \frac{dA}{d\theta }=4R^2\left[-{\sin }^4\theta +3{\sin }^2\theta {\cos }^2\theta \right] \\ \Rightarrow \frac{dA}{d\theta }=4R^2{\sin }^2\theta \left[3{\cos }^2\theta -{\sin }^2\theta \right] \end{gathered}$$

Put $\frac{dA}{d\theta }=0$ to find the critical points.

$$\begin{gathered} 4R^2{\sin }^2\theta \left[3{\cos }^2\theta -{\sin }^2\theta \right]=0 \\ \Rightarrow 3{\cos }^2\theta -{\sin }^2\theta =0 \\ \Rightarrow 3{\cos }^2\theta ={\sin }^2\theta \\ \Rightarrow 3={\tan }^2\theta \\ \Rightarrow \tan \theta =\sqrt{3} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$

So, the area of the trapezium is maximum when $\theta =\frac{\mathrm{\pi }}{3}$.

So,

$$\begin{gathered} A=4R^2\cos \left(\frac{\mathrm{\pi }}{3}\right){\sin }^3\left(\frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow A=4R^2\left(\frac{1}{2}\right){\left(\frac{\sqrt{3}}{2}\right)}^3 \\ \Rightarrow A=\frac{3\sqrt{3}{R}^2}{4} \end{gathered}$$

Therefore, the largest area of a trapezium inscribed in a semicircle of radius $R$, if the lower base is on the diameter, is $\frac{3\sqrt{3}{R}^2}{4}$.

Hence, option(A) is the correct answer i.e. $\frac{3\sqrt{3}{R}^2}{4}$